Question
Equal volumes of solutions in two different vessels are represented above. If the solution represented in vessel 1 is KCl(aq), then them solution represented in vessel 2 could be an aqueous solution of
(A) KCl with the same molarity as the solution in vessel 1
(B) KCl with twice the molarity of the solution in vessel 1
(C) \( CaCl_{2}\) with the same molarity as the solution in vessel 1
(D) \(CaCl_{2}\) with twice the molarity of the solution in vessel 1
▶️Answer/Explanation
Ans:D
Based on the image, Vessel 1 contains an equal number of positive and negative ions, indicating a 1:1 electrolyte like KCl. In Vessel 2, there are more positive ions than negative ions, suggesting a higher ratio of positive to negative ions compared to Vessel 1.
Therefore, the correct option is (D) CaCl₂ with twice the molarity of the solution in Vessel 1.
The higher ratio of positive to negative ions in Vessel 2 is consistent with the presence of CaCl₂, which has a 2:1 ratio of positive to negative ions (Ca²⁺ and 2Cl⁻). Additionally, the higher number of ions overall in Vessel 2 compared to Vessel 1 suggests a higher molarity (twice the molarity of Vessel 1’s KCl solution).
Question
What volume of a 0.100M HCl stock solution should be used to prepare 250.00mL of 0.0250M HCl?
▶️Answer/Explanation
Ans: C
When using a stock solution to prepare a solution of lower concentration, only solvent is added and the number of moles of solute remain constant. In these situations, the volume of stock solution can be obtained by rearranging the equation: \(M_1V_1=M_2V_2\). Therefore, \(V_1=\frac{M_2V_2}{M_1}\)=0.0250M×250.00mL0.100M=62.5mL. Note that when working with ratios, the units of volume will be the same as those of the volume given.
Question
A 500.mL aqueous solution of \(Na_3PO_4\) (molarmass=164g/mol) was prepared using 82g of the solute. What is the molarity of \(Na_3PO_4\) in the resulting solution?
A 0.0010M
B 0.16M
C 0.25M
D 1.0M
▶️Answer/Explanation
Ans: D
To calculate the molarity of a solution, the number of moles of solute is divided by the volume of the solution in liters. Therefore, \( \frac{82/164}{500/1000}=1.0\) M
Question
\(C_3H_8(g) + 4 Cl_2(g) → C_3H_4Cl(g) + 4 HCl(g)\)
A 6.0 mol sample of\( C_{3}H_8(g)\) and a 20. mol sample of \(Cl_{2}\)(g) are placed in a previously evacuated vessel, where they react according to the equation above. After one of the reactants has been totally consumed, how many moles of HCl(g) have been produced?
(A) 4.0 mo
(B) 8.0 mol
(C) 20. mol
(D) 24 mol
▶️Answer/Explanation
Ans:C